Question: Is ${116585}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {116585}= &&{1}\cdot100000+ \\&&{1}\cdot10000+ \\&&{6}\cdot1000+ \\&&{5}\cdot100+ \\&&{8}\cdot10+ \\&&{5}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {116585}= &&{1}(99999+1)+ \\&&{1}(9999+1)+ \\&&{6}(999+1)+ \\&&{5}(99+1)+ \\&&{8}(9+1)+ \\&&{5} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {116585}= &&\gray{1\cdot99999}+ \\&&\gray{1\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {1}+{1}+{6}+{5}+{8}+{5} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${116585}$ is divisible by $3$ if ${ 1}+{1}+{6}+{5}+{8}+{5}$ is divisible by $3$ Add the digits of ${116585}$ $ {1}+{1}+{6}+{5}+{8}+{5} = {26} $ If ${26}$ is divisible by $3$ , then ${116585}$ must also be divisible by $3$ ${26}$ is not divisible by $3$, therefore ${116585}$ must not be divisible by $3$.